Find the value of $x$ from the equation:$$\log_{(2x+5)}{(10x^2+29x+10)}=5-\log_{(5x+2)}{(4x^2+20x+25)}$$
This question seems bit tricky (especially because the bases of the log are different) and on first instance leaves clueless on how to proceed. How to find value of $x$ from this equation.
$$\log _{ (2x+5) }{ (10x^{ 2 }+29x+10) } +\log _{ (5x+2) }{ (4x^{ 2 }+20x+25) } =5\\ \log _{ \left( 2x+5 \right) }{ \left( 2x+5 \right) \left( 5x+2 \right) } +\log _{ \left( 5x+2 \right) }{ { \left( 2x+5 \right) }^{ 2 } } =5\\ \log _{ \left( 2x+5 \right) }{ \left( 2x+5 \right) +\log _{ \left( 2x+5 \right) }{ \left( 5x+2 \right) } +2\log _{ \left( 5x+2 \right) }{ \left( 2x+5 \right) } =5 } \\ 1+\log _{ \left( 2x+5 \right) }{ \left( 5x+2 \right) } +2\log _{ \left( 5x+2 \right) }{ \left( 2x+5 \right) } =5\\ \log _{ \left( 2x+5 \right) }{ \left( 5x+2 \right) } +2\log _{ \left( 5x+2 \right) }{ \left( 2x+5 \right) } =4$$ now solve quadratic equation and set$$\log _{ \left( 2x+5 \right) }{ \left( 5x+2 \right) } =y\\ y+\frac { 2 }{ y } =4$$
The answer is the following:
$x = -5/2$
Here's how to derive it:
Use the change of base formula to make everything in terms of $\log_{5 x + 2}$:
${ \log_{5x+2} (10 x^2 + 29 x + 10) \over \log_{5x+2}(2 x + 5)} = 5 - \log_{5 x+2} (4 x^2 + 20 x + 25)$
and noting $5 = \log_{5 x +2} (5 x +2)^5$,
then
$(10 x^2 + 29 x + 10) = (2 x+5)((5x+2)^5-(4 x^2 + 20 x + 25))$
Which gives the solution quoted above.
you can write $$\frac{\ln(10x^2+20x+10)}{\ln(2x+5)}=5-\frac{\ln(4x^2+20x+25)}{\ln(5x+2)}$$ note that $$10x^2+20x+10=10(x+1)^2$$ and $$4x^2+20x+25=(2x+5)^2$$
Hint:
Note that
$$ 10x^2 +29x+10= (5x+2)(2x+5) $$
and $$ 4x^2 +20x+25=(2x+5)^2 $$
and remember that $$log_a b=\frac{1}{\log_b a}$$
